Let $F_n$ be the free group on $n$ letters. Let $g_1,...,g_{2m} \in F_n$, can the group $$F_n / \langle\langle[g_1,g_2],...,[g_{2m-1},g_{2m}]\rangle\rangle$$ ever have torsion elements?
The double angle brackets means "normal subgroup generated by" and $[a,b] = aba^{-1}b^{-1}$.
This problem arose when I saw a question in an old paper of Yanagawa ("On Ribbon 2-knots, II") that mentioned that it was unknown if a complement of a ribbon 2-knot could have torsion. These groups have presentations that are special cases of what I asked about in the question so I was guessing there was a simple counterexample to my question, but I had no inspiration to find it. Just FYI - the main result of that paper is a well known open problem (are ribbon disk complements aspherical?), so the proof of the main result is flawed... Although I didn't go looking for the error.
Edit : I just wanted to clarify that the elements $g_i$ are arbitrary elements (not necessarily the generators). So user1729's nice answer answers the question in the case where the $g_i$ are generators, but I am still interested in the question for general $g_i$. Also, for those interested in the topological origin of this problem, I actually messed up with the relations that arise in the context of the aforementioned paper. The groups that arise as ribbon group complements are of the form $F_n / << x_1 = x_2^{g_1}, x_2 = x_3^{g_2},...,x_{n_1} = x_n^{g_n} >>$ where here the $x_i$ are the generators of $F_n$ and again the $g_i$ are arbitrary elements of $F_n$. I would like to know if these groups can have any torsion as well.
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